Problem: Let $g$ be a vector-valued function defined by $g(t)=\left(4\log(t),\dfrac{4}{t+2}\right)$. Find $g'(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(\dfrac{4}{t\ln(10)},-\dfrac{4}{(t+2)^2}\right)$ (Choice B) B $\left(\dfrac{4}{t},\dfrac{8}{(t+2)^2}\right)$ (Choice C) C $\left(\dfrac{1}{t\ln(4)},\dfrac{4}{(t+2)^2}\right)$ (Choice D) D $\dfrac{4}{t}-\dfrac{4}{(t+2)^2}$
Solution: $g$ is a vector-valued function. This means it takes one number as an input $(t)$, but it outputs two numbers as a two-dimensional vector. Finding the derivative of a vector-valued function is pretty straightforward. Suppose a vector-valued function is defined as $u(t)=(v(t),w(t))$, then its derivative is the vector-valued function $u'(t)=(v'(t),w'(t))$. In other words, the derivative is found by differentiating each of the expressions in the function's output vector. Recall that $g(t)=\left(4\log(t),\dfrac{4}{t+2}\right)$. Let's differentiate the first expression: $\dfrac{d}{dt}[4\log(t)]=\dfrac{4}{t\ln(10)}$ Let's differentiate the second expression: $\dfrac{d}{dt}\left(\dfrac{4}{t+2}\right)=-\dfrac{4}{(t+2)^2}$ Now let's put everything together: $\begin{aligned} g'(t)&=\left(\dfrac{d}{dt}[4\log(t)],\dfrac{d}{dt}\left(\dfrac{4}{t+2}\right)\right) \\\\ &=\left(\dfrac{4}{t\ln(10)},-\dfrac{4}{(t+2)^2}\right) \end{aligned}$ In conclusion, $g'(t)=\left(\dfrac{4}{t\ln(10)},-\dfrac{4}{(t+2)^2}\right)$.